Past exam of the mathematics course of the University of Cambridge / 2001 / ib / Paper 2 / 10A / Solution

A metric space is totally bounded if, for every $ϵ>0$, finitely many open balls of radius $ϵ$ cover it. The Bolzano-Weierstrass property here means that every sequence has a subsequence converging to a point of the space, namely sequential compactness.

First suppose that property holds. A Cauchy sequence has a convergent subsequence with limit $x$ in the space. Given $ϵ>0$, choose an index after which all sequence terms lie within $ϵ/2$ of each other, and a subsequence term beyond that index within $ϵ/2$ of $x$. The triangle inequality then puts every subsequent term within $ϵ$ of $x$. Thus the whole Cauchy sequence converges, proving completeness.

If total boundedness failed, some $ϵ>0$ would admit no finite ball cover. Choose points inductively outside the union of the $ϵ$-balls about previously chosen points. Distinct sequence terms then have distance at least $ϵ$. No subsequence can be Cauchy, and therefore none can converge, a contradiction. This proves total boundedness.

Conversely, suppose the space is complete and totally bounded. For any sequence, cover the space by finitely many radius-$2^{-1}$ balls and retain the infinitely many indices in one ball. Within those indices, a finite cover by radius-$2^{-2}$ balls retains an infinite subset in one smaller ball. Repeat with radii $2^{-k}$ to obtain nested infinite index sets $I_k$. Choose increasing indices $n_k\in I_k$. For $l,j\ge k$, both terms lie in the ball selected at step $k$, soThe subsequence is Cauchy and completeness makes it converge within the space. Hence the Bolzano-Weierstrass property is equivalent to completeness plus total boundedness.